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If log⁡ab−c=log⁡bc−a=log⁡ca−b\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}, prove that aa×bb×cc=1a^{a}\times b^{b}\times c^{c}=1

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Full solution

Q. If log⁡ab−c=log⁡bc−a=log⁡ca−b\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}, prove that aa×bb×cc=1a^{a}\times b^{b}\times c^{c}=1
  1. Equating Logarithmic Arguments: Since log⁡ab−c=log⁡bc−a=log⁡ca−b\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}, we can equate the arguments of the logarithms:\newlineab−c=bc−a=ca−b \frac{a}{b-c} = \frac{b}{c-a} = \frac{c}{a-b}
  2. Cross-Multiply Relationships: Cross-multiply each pair to find relationships between aa, bb, and cc:\newlinea(c−a)=b(b−c)andb(a−b)=c(c−a) a(c-a) = b(b-c) \quad \text{and} \quad b(a-b) = c(c-a)
  3. Simplify and Rearrange: Simplify and rearrange the equations:\newlineac−a2=b2−bcandab−b2=c2−ca ac - a^2 = b^2 - bc \quad \text{and} \quad ab - b^2 = c^2 - ca
  4. Addition of Equations: Add the equations together:\newlineac−a2+ab−b2=b2−bc+c2−ca ac - a^2 + ab - b^2 = b^2 - bc + c^2 - ca \newlineThis simplifies to:\newlineac+ab−a2−b2=b2+c2−bc−ca ac + ab - a^2 - b^2 = b^2 + c^2 - bc - ca
  5. Isolate Variables: Rearrange terms to isolate all variables on one side:\newlineac+ab+bc=a2+b2+c2 ac + ab + bc = a^2 + b^2 + c^2
  6. Exponential Transformation: Taking exponentials on both sides, we get:\newlineeac+ab+bc=ea2+b2+c2 e^{ac + ab + bc} = e^{a^2 + b^2 + c^2}
  7. Incorrect Exponential Form: Using properties of exponents, rewrite the equation:\newline(aa×bb×cc)2=ea2+b2+c2 (a^a \times b^b \times c^c)^2 = e^{a^2 + b^2 + c^2} \newlineThis step is incorrect as the exponential form does not match the original equation.

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